Pushing force making an angle $θ$ to the horizontal is applied on a block of weight w placed on a horizontal table. If the angle of friction is $ϕ$ , the magnitude of force required to move the body is equal to

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$\frac{w \sin ϕ}{\tan (θ - ϕ)}$

$\frac{w \tan ϕ}{\sin (θ - ϕ)}$

$\frac{w \cos ϕ}{\cos (θ - ϕ)}$

$\frac{w \sin ϕ}{\cos (θ + ϕ)}$

answer is B.

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Detailed Solution

$N = w + F \sin θ$

$∴ f_{\max} = μ N = (\tan ϕ) (w + F \sin θ)$

To move the body, $F \cos θ = f_{\max} = (\tan ϕ) (w + F \sin θ)$

Solving this equation, we get

$F = \frac{w \sin ϕ}{\cos (θ + ϕ)}$

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